Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 38 / 6 / c / Solution

Row operations multiply each original tableau equation by an invertible matrix. The slack variable block therefore records that matrix and allows the original payoff matrix to be recovered without guessing.

From the first tableau, let $X$ be the coefficient block of $x$ and $S$ that of $s$. Since its original equations were $Q^{T}x+s=1$, we have $Q^T=S^{-1}X$. HeresoFor the second tableau, writing its $y$ and $r$ blocks as $Y,R$, the original equations $Py+r=1$ give $P=R^{-1}Y$. We obtainThus a representative pair of payoff matrices isBoth reconstructed right-hand sides are $1$, as a check on the tableau normalization. Independent transformations $P\mapsto aP+b11^T$, $Q\mapsto cQ+d11^T$, with $a,c>0$, preserve best responses and hence identify the same strategic solution up to positive affine payoff transformations.

Directly,The supports of $x^{\prime }$ and $y^{\prime }$ lie entirely among their respective best-response coordinates, confirming the Nash equilibrium found above. Since this representative is a symmetric bimatrix game, swapping the players' strategies preserves the Nash equilibrium conditions. Explicitly, $P{x}^{\prime }=(3,2,3{)}^T$ is maximized on the support of $y^{\prime }$, and $Q^T{y}^{\prime }=P{y}^{\prime }=(1,3,3{)}^T$ is maximized on the support of $x^{\prime }$. Therefore